Dixon's Hypergeometric Theorem/Examples/3F2(0.5,0.5,0.5;1,1;1)

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Example of Use of Dixon's Hypergeometric Theorem

$1 + \paren {\dfrac 1 2}^3 + \paren {\dfrac {1 \times 3} {2 \times 4} }^3 + \paren {\dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} }^3 + \cdots = \dfrac \pi {\paren {\map \Gamma {\dfrac 3 4} }^4 }$


Proof

From Dixon's Hypergeometric Theorem:

$\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {\dfrac n 2 + 1} \map \Gamma {x + y + \dfrac n 2 + 1} } { \map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {y + \dfrac n 2 + 1} } $

where:

$\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1}$ is the generalized hypergeometric function of $1$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { n^{\overline k} \paren {-x}^{\overline k} \paren {-y}^{\overline k} } { \paren {x + n + 1}^{\overline k} \paren {y + n+ 1}^{\overline k} } \dfrac {1^k} {k!}$
$x^{\overline k}$ denotes the $k$th rising factorial power of $x$
$\map \Gamma {n + 1} = n!$ is the Gamma function.


We have:

\(\ds \map { {}_3 \operatorname F_2} { { {\dfrac 1 2, \dfrac 1 2, \dfrac 1 2} \atop {1, 1} } \, \middle \vert \, 1}\) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \dfrac { \paren {\dfrac 1 2}^{\overline k} \paren {\dfrac 1 2}^{\overline k} \paren {\dfrac 1 2}^{\overline k} } { 1^{\overline k} 1^{\overline k} } \dfrac {1^k} {k!}\) Definition of Generalized Hypergeometric Function
\(\ds \) \(=\) \(\ds 1 + \dfrac {\paren {\dfrac 1 2}^3} {\paren {1!}^3} + \dfrac {\paren {\dfrac 1 2 \times \dfrac 3 2}^3} {\paren {2!}^3} + \dfrac {\paren {\dfrac 1 2 \times \dfrac 3 2 \times \dfrac 5 2 }^3} {\paren {3!}^3} + \cdots\) One to Integer Rising is Integer Factorial, $1^k = 1$, Number to Power of Zero Rising is One
\(\ds \) \(=\) \(\ds 1 + \paren {\dfrac 1 2}^3 + \paren {\dfrac {1 \times 3} {2 \times 4} }^3 + \paren {\dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} }^3 + \cdots\)

and:

\(\ds \map { {}_3 \operatorname F_2} { { {\dfrac 1 2, \dfrac 1 2, \dfrac 1 2} \atop {1, 1} } \, \middle \vert \, 1}\) \(=\) \(\ds \dfrac {\map \Gamma {-\dfrac 1 2 + \dfrac 1 2 + 1} \map \Gamma {-\dfrac 1 2 + \dfrac 1 2 + 1} \map \Gamma {\dfrac {\dfrac 1 2} 2 + 1} \map \Gamma {-\dfrac 1 2 -\dfrac 1 2 + \dfrac {\dfrac 1 2} 2 + 1} } { \map \Gamma {\dfrac 1 2 + 1} \map \Gamma {-\dfrac 1 2 - \dfrac 1 2 + \dfrac 1 2 + 1} \map \Gamma {-\dfrac 1 2 + \dfrac {\dfrac 1 2} 2 + 1} \map \Gamma {-\dfrac 1 2 + \dfrac {\dfrac 1 2} 2 + 1} }\) Dixon's Hypergeometric Theorem
\(\ds \) \(=\) \(\ds \dfrac {\map \Gamma {1} \map \Gamma 1 \map \Gamma {\dfrac 5 4} \map \Gamma {\dfrac 1 4 } } { \map \Gamma {\dfrac 3 2} \map \Gamma {\dfrac 1 2 } \map \Gamma {\dfrac 3 4} \map \Gamma {\dfrac 3 4} }\) simplifying
\(\ds \) \(=\) \(\ds \dfrac {\dfrac 1 4 \map \Gamma {\dfrac 1 4} \map \Gamma {\dfrac 1 4} } { \dfrac 1 2 \map \Gamma {\dfrac 1 2} \map \Gamma {\dfrac 1 2 } \map \Gamma {\dfrac 3 4} \map \Gamma {\dfrac 3 4} }\) Definition of Gamma Function and $\map \Gamma {1} = 1$
\(\ds \) \(=\) \(\ds \dfrac {\paren {\map \Gamma {\dfrac 1 4} }^2} {2 \pi \paren {\map \Gamma {\dfrac 3 4 } }^2}\) Gamma Function of One Half


Recall from the Euler's Reflection Formula: $\map \Gamma z \map \Gamma {1 - z} = \dfrac \pi {\map \sin {\pi z} }$

Therefore:

\(\ds \map \Gamma {\dfrac 1 4} \map \Gamma {1 - \dfrac 1 4}\) \(=\) \(\ds \dfrac \pi {\map \sin {\dfrac \pi 4} }\)
\(\ds \map \Gamma {\dfrac 1 4} \map \Gamma {\dfrac 3 4}\) \(=\) \(\ds \dfrac \pi {\dfrac {\sqrt 2} 2}\) Sine of $45 \degrees$
\(\ds \map \Gamma {\dfrac 1 4}\) \(=\) \(\ds \dfrac {\sqrt 2 \pi} {\map \Gamma {\dfrac 3 4} }\)


Substituting this result back into our equation above:

\(\ds \map { {}_3 \operatorname F_2} { { {\dfrac 1 2, \dfrac 1 2, \dfrac 1 2} \atop {1, 1} } \, \middle \vert \, 1}\) \(=\) \(\ds \dfrac {\paren {\map \Gamma {\dfrac 1 4} }^2} {2 \pi \paren {\map \Gamma {\dfrac 3 4 } }^2}\) from above
\(\ds \) \(=\) \(\ds \dfrac {\paren {\dfrac {\sqrt 2 \pi} {\map \Gamma {\dfrac 3 4} } }^2 } { 2 \pi \paren {\map \Gamma {\dfrac 3 4} }^2}\)
\(\ds \) \(=\) \(\ds \dfrac \pi {\paren {\map \Gamma {\dfrac 3 4} }^4}\)


Therefore:

$1 + \paren {\dfrac 1 2}^3 + \paren {\dfrac {1 \times 3} {2 \times 4} }^3 + \paren {\dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} }^3 + \cdots = \dfrac \pi {\paren {\map \Gamma {\dfrac 3 4} }^4}$

$\blacksquare$


Sources

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